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Represent the given irrational numbers $\sqrt{3.5}, \sqrt{9.4}, \sqrt{10.5}$ on the real number line, and choose which number comes middle of the line.

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\sqrt{3.5}

\sqrt{9.4}

\sqrt{10.5}

None

answer is B.

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Detailed Solution

(i) The first number we have =

\sqrt{3.5}

Now we can write that,

= \sqrt{{(\frac{3.5 + 1}{2})}^{2} - {(\frac{3.5 - 1}{2})}^{2}}

= \sqrt{{(\frac{4.5}{2})}^{2} - {(\frac{2.5}{2})}^{2}}

= (2.25)^{2} - (1.5)^{2}

Now, to draw given number of number lien make a line

BC with measure 1.25 cm

From point C, make a ray CX with the angle measure of

90^{\circ}

From point B as center and 2.25 cm radius, make an arc that intersects CX at the point A. Join A to C.
Length of this AC using Pythagoras theorem,

AC = \sqrt{A B^{2} - B C^{2}}

= \sqrt{(2.25)^{2} - (1.25)^{2}}

= \sqrt{5.0625 - 1.5625}

= \sqrt{3.5}

(ii) Number given in second part =

\sqrt{9.4}

Here,

= \sqrt{{(\frac{9.4 + 1}{2})}^{2} - {(\frac{9.4 - 1}{2})}^{2}}

= \sqrt{{(\frac{10.4}{2})}^{2} - {(\frac{8.4}{2})}^{2}} = \sqrt{(5.2)^{2} - (4.2)^{2}}

Make BC = 4.2 cm. From point C, make ray CX with angle measure of

90^{\circ}

From B as centre with 5.2 cm radius make an arc that intersects CX at point A. Join A to C. Length of AC,

AC =

= \sqrt{{AB}^{2} - {BC}^{2}}

= \sqrt{(5.2)^{2} - (4.2)^{2}}

= \sqrt{27.04 - 17.64}

= \sqrt{9.4}

(iii) Here, we have,

\sqrt{10.5}

Now,

= \sqrt{{(\frac{10.5 + 1}{2})}^{2} - {(\frac{10.5 - 1}{2})}^{2}}

= \sqrt{{(\frac{11.5}{2})}^{2} - {(\frac{9.5}{2})}^{2}} = \sqrt{(5.75)^{2} - (4.75)^{2}}

In the same way as perviours the line segment AC,

AC = \sqrt{(5.75)^{2} - (4.75)^{2}}

= \sqrt{33.0625 - 22.5625}

= \sqrt{10.5000} = \sqrt{10.5}

Option 2 is correct.

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